At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure…
At $10^{\circ} \mathrm{C}$ the value of the density of a fixed mass of an ideal gas divided by its pressure is $\mathrm{x}$. At $110^{\circ} \mathrm{C}$ this ratio is
$\frac{283}{383} x$
$\mathrm{x}$
$\frac{383}{283} x$
$\frac{10}{110} x$
Solution
$\begin{aligned}
P V & =n R T \\
& \Rightarrow P \cdot \frac{m}{\rho}=\frac{m}{M} R T \\
& \Rightarrow \frac{\left(\frac{\rho}{P}\right)_f}{\left(\frac{\rho}{P}\right)_i}=\frac{T_i}{T_f}=\frac{10+273}{110+273}=\frac{283}{383} \\
& \Rightarrow \frac{\left(\frac{\rho}{P}\right)_f}{x}=\frac{283}{383} \\
& \therefore\left(\frac{\rho}{P}\right)_f=\frac{283}{383} x
\end{aligned}$
^